(*^ ::[ Information = "This is a Mathematica Notebook file. It contains ASCII text, and can be transferred by email, ftp, or other text-file transfer utility. It should be read or edited using a copy of Mathematica or MathReader. If you received this as email, use your mail application or copy/paste to save everything from the line containing (*^ down to the line containing ^*) into a plain text file. On some systems you may have to give the file a name ending with ".ma" to allow Mathematica to recognize it as a Notebook. The line below identifies what version of Mathematica created this file, but it can be opened using any other version as well."; FrontEndVersion = "NeXT Mathematica Notebook Front End Version 2.2"; NeXTStandardFontEncoding; fontset = title, inactive, noPageBreakBelow, noPageBreakInGroup, nohscroll, preserveAspect, groupLikeTitle, center, M7, bold, L1, e8, 24, "Times"; ; fontset = subtitle, inactive, noPageBreakBelow, noPageBreakInGroup, nohscroll, preserveAspect, groupLikeTitle, center, M7, bold, L1, e6, 18, "Times"; ; fontset = subsubtitle, inactive, noPageBreakBelow, noPageBreakInGroup, nohscroll, preserveAspect, groupLikeTitle, center, M7, italic, L1, e6, 14, "Times"; ; fontset = section, inactive, noPageBreakBelow, nohscroll, preserveAspect, groupLikeSection, grayBox, M22, bold, L1, a20, 18, "Times"; ; fontset = subsection, inactive, noPageBreakBelow, nohscroll, preserveAspect, groupLikeSection, blackBox, M19, bold, L1, a15, 14, "Times"; ; fontset = subsubsection, inactive, noPageBreakBelow, nohscroll, preserveAspect, groupLikeSection, whiteBox, M18, bold, L1, a12, 12, "Times"; ; fontset = text, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = smalltext, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 10, "Times"; ; fontset = input, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeInput, M42, N23, bold, L1, 12, "Courier"; ; fontset = output, output, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeOutput, M42, N23, L-5, 12, "Courier"; ; fontset = message, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeOutput, M42, N23, L1, 12, "Courier"; ; fontset = print, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeOutput, M42, N23, L1, 12, "Courier"; ; fontset = info, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeOutput, M42, N23, L1, 12, "Courier"; ; fontset = postscript, PostScript, formatAsPostScript, output, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeGraphics, M7, l34, w282, h287, L1, 12, "Courier"; ; fontset = name, inactive, noPageBreakInGroup, nohscroll, preserveAspect, M7, italic, B65535, L1, 10, "Times"; ; fontset = header, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, italic, L1, 12, "Times"; ; fontset = leftheader, 12; fontset = footer, inactive, nohscroll, noKeepOnOnePage, preserveAspect, center, M7, italic, L1, 12, "Times"; ; fontset = leftfooter, 12; fontset = help, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = clipboard, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = completions, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12, "Courier"; ; fontset = special1, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = special2, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = special3, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = special4, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = special5, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; paletteColors = 128; automaticGrouping; magnification = 150; currentKernel; ] :[font = title; inactive; preserveAspect; startGroup] Kinematic Analysis of a Reciprocating Engine :[font = section; inactive; preserveAspect; startGroup] BRIEF ABSTRACT :[font = subsection; inactive; preserveAspect; endGroup] With the given realistic data, the main goal is to determine the piston's position with respect to time as well as the connecting rod's angle from the piston axis with respect to time. This problem is designed to be completed within a single class period once students are familiar with the concepts of position, velocity and acceleration. :[font = section; inactive; Cclosed; preserveAspect; startGroup] GENERAL INFORMATION :[font = subsection; inactive; preserveAspect] FileName: ENGINE :[font = subsection; inactive; preserveAspect] Full title: Kinematic Analysis of a Reciprocating Engine :[font = subsection; inactive; preserveAspect] Last Update: 5/30/96 :[font = subsection; inactive; preserveAspect] Developer: Jerry Fine Department of Mechanical Engineering Rose-Hulman Institute of Technology Terre Haute, IN 47803 Phone: 812-877-8353. Email: fine@nextwork.rose-hulman.edu FAX: 812-877-3198 :[font = subsection; inactive; preserveAspect] Contact: Aaron Klebanoff, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8151. Email: Klebanoff@rose-hulman.edu. FAX: 812-877-3198. :[font = subsection; inactive; preserveAspect; endGroup] Support: The production of this material is supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849: Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering and the Arvin Foundation of Columbus IN. :[font = section; inactive; Cclosed; preserveAspect; startGroup] STATEMENT OF PROBLEM :[font = text; inactive; preserveAspect] The reciprocating engine plays an important part in modern life. You may be interested in studying the mathematics behind this device, and this problem provides an opportunity for you to do that. 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FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF0C3FFF FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF grestore end %%EndFile grestore %End of Graphics MathPictureEnd :[font = text; inactive; preserveAspect] As the crankshaft rotates, the piston moves back and forth along the x-axis. Your job is to find the position, velocity and acceleration of the piston as a function of the angle q that the crank makes with the x-axis. Since the crankshaft is assumed to be rotating at a constant angular velocity w, (measured in radians/sec), this is equivalent to finding the quantities as functions of time. (q = w t) You are also required to find the angular velocity and angular acceleration of the connecting rod. These will be the first and second derivatives with respect to time of the angle f that the connecting rod makes with the x-axis. The quantities x, dx/dt, d2x/dt2, f, df/dt d2f/dt2 should be plotted as either functions of time, or as functions of q. Find the maximum and minimum values of each quantity. The maximum and minimum values are of interests to engineers calculating the stresses in the engine. ;[s] 24:0,0;182,1;183,2;302,3;303,4;402,5;403,6;406,7;407,8;710,9;711,10;715,11;716,12;730,13;732,14;748,15;749,16;767,17;768,18;769,19;772,20;773,21;840,22;841,23;999,-1; 24:1,11,8,Times,0,12,0,0,0;1,0,0,Symbol,0,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,0,0,Symbol,0,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,0,0,Symbol,0,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,0,0,Symbol,0,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,11,8,Times,32,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,11,8,Times,32,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,0,0,Symbol,0,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,0,0,Symbol,0,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,11,8,Times,32,12,0,0,0;1,0,0,Symbol,0,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,11,8,Times,32,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,0,0,Symbol,0,12,0,0,0;1,11,8,Times,0,12,0,0,0; :[font = subsection; inactive; preserveAspect; startGroup] Data :[font = text; inactive; preserveAspect] Here is some realistic data that you can use. Let the crank length, r, be 8 cm, and the connecting rod length be 24.5 cm. Assume that the crank is turning at a constant 1000 rpm, (revolutions per minute). We must convert this to radians per second. Let T be the time in seconds it takes to do one rotation. :[font = input; preserveAspect; endGroup; endGroup] data = { r -> .08, L -> .245, w -> 1000 (2 Pi / 60)} //N; T = N[60/1000]; :[font = section; inactive; Cclosed; preserveAspect; startGroup] KEYWORDS :[font = text; inactive; preserveAspect; endGroup] Piston, crank, connecting rod, kinematics, position, velocity and acceleration. :[font = section; inactive; Cclosed; preserveAspect; startGroup] TEACHER NOTES :[font = subsection; inactive; preserveAspect; startGroup] ISSUES RELATED TO THE PROBLEM :[font = text; inactive; preserveAspect; endGroup] There is often a lot of interest in how the motor in a car works. This problem is oriented towards the physics of the piston-cylinder arrangement. It suggests the first step in finding the forces and torques in the engine, this step being an analysis of the engine's motion. The first goals, then, are the acceleration of the piston, as well as the angular velocity and acceleration of the connecting rod. The accelerations of the centers of mass of the connecting rod and the crank are also important; they may be investigated in an extension to this problem. The kinematics of bodies which move in a plane is not too difficult. In this case it is possible to use the geometry of a simple triangle which represents the instantaneous configuration of the device. Two of the sides of the triangle have constant length. The independent variable is q, the angle the crank makes with the positive x direction. Differentiation of the trigonometric description of the triangle produces the needed expressions. One difficulty that the students face is the management of the unknown quantities in the problem. It requires some definite advance planning. If the student is using a CAS, then this issue is less important. ;[s] 3:0,0;863,1;864,2;1235,-1; 3:1,11,8,Times,0,12,0,0,0;1,0,0,Symbol,0,12,0,0,0;1,11,8,Times,0,12,0,0,0; :[font = subsection; inactive; preserveAspect; startGroup] Prerequisites :[font = text; inactive; preserveAspect; endGroup] The student should be aware of the definitions of position, velocity and acceleration. The student should have some familiarity with geometry, and be able to differentiate using the chain rule. The student must also have enough familiarity with reciprocating engines to be able to visualize the motion. This may be a problem for some students. Many science books contain good 'cut-away' drawings of the engine. It might be a good idea to have the student sketch the basic configuration for several different crank angles, including 0, 90, 180 and 270 degrees. :[font = subsection; inactive; preserveAspect; startGroup] Time allotment - time management :[font = subsubsection; inactive; preserveAspect; startGroup] If you are using a computer algebra system :[font = text; inactive; preserveAspect; endGroup] The problem is not very hard. The key is to identify the necessary geometric relationships (see POSSIBLE SOLUTIONS) and then to key them into the system. The plots may then be obtained posthaste. It would probably be a 20 minute problem. There would be some time to explore the suggested extensions. :[font = subsubsection; inactive; preserveAspect; startGroup] If you are not using a computer algebra solution :[font = text; inactive; preserveAspect; endGroup; endGroup] This problem is a lengthy one. It takes students some time just to grasp what needs to be done. The work of grinding out the derivatives and seeing how the pieces fit together to give the solution is also long. This problem might take more than 50 minutes unless hints were given. If you give hints and stop at finding the velocity and acceleration of the piston, students might finish in less than 50 minutes. The use of a graphing calculator is strongly recommended. :[font = subsection; inactive; preserveAspect; startGroup] Expectations :[font = text; inactive; preserveAspect; endGroup] Trouble may result from the following: (1) Not knowing how to use the chain rule in this situation. (2) Defeat due to complexity of expressions. If the problem is done by hand, and the plots done with a graphing calculator, then develop the expressions in the following order: f, df/dt d2f/dt2 x, dx/dt, d2x/dt2 Do not substitute these expressions into each other. The expressions just get too messy! If you proceed in this order and enter the functions into the graphing calculator correctly, then things should work out pretty well. ;[s] 14:0,0;293,1;295,2;311,3;312,4;330,5;331,6;332,7;335,8;336,9;387,10;388,11;392,12;393,13;620,-1; 14:1,11,8,Times,0,12,0,0,0;1,0,0,Symbol,0,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,0,0,Symbol,0,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,11,8,Times,32,12,0,0,0;1,0,0,Symbol,0,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,11,8,Times,32,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,11,8,Times,32,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,11,8,Times,32,12,0,0,0;1,11,8,Times,0,12,0,0,0; :[font = subsection; inactive; preserveAspect; startGroup] Future payoffs :[font = text; inactive; preserveAspect; endGroup] This problem is intended to help students begin to recognize the links between elementary calculus and kinematics. It connects these ideas to the real world. :[font = subsection; inactive; preserveAspect; startGroup] Extensions :[font = text; inactive; preserveAspect; endGroup] There is an obvious extension to the problem: the dynamic analysis of the engine. To do the above extension, one needs the velocity and acceleration of the mass centers of the crank and the connecting rod, in addition to those of the piston, which has just been computed. :[font = subsection; inactive; preserveAspect; endGroup] Reference: Design of Machinery, R.L. Norton, McGraw Hill, 1992. ;[s] 2:0,0;11,1;64,-1; 2:1,12,9,Times,1,14,0,0,0;1,11,8,Times,0,12,0,0,0; :[font = section; inactive; Cclosed; preserveAspect; startGroup] POSSIBLE SOLUTION :[font = text; inactive; preserveAspect] The solutions may be developed from the geometry of this triangle. The two relationships that get you to the solution are: x = r cos(q) + L cos(f) r sin(q) = L sin(f) ;[s] 9:0,0;144,1;145,2;155,3;156,4;172,5;173,6;183,7;184,8;186,-1; 9:1,11,8,Times,0,12,0,0,0;1,0,0,Symbol,0,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,0,0,Symbol,0,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,0,0,Symbol,0,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,0,0,Symbol,0,12,0,0,0;1,11,8,Times,0,12,0,0,0; :[font = input; preserveAspect] data = { r -> .08, L -> .245, w -> 1000 (2 Pi / 60)} //N; T = N[60/1000]; :[font = text; inactive; preserveAspect] It is very easy to implement these if you are using a computer algebra system such as Mathematica. :[font = input; preserveAspect] x[t_] := r Cos[w t] - L Cos[ fi[t] ] /. data :[font = input; preserveAspect] fi[t_] := Pi - ArcSin[ r / L Sin[w t] ] /. data :[font = subsection; inactive; preserveAspect; startGroup] Plots of Desired Functions :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] Connecting Rod Angle :[font = input; preserveAspect; startGroup] Plot[fi[t]/Degree, {t, 0, T}, PlotLabel -> "Connecting Rod Angular Position", AxesLabel -> {"t [sec]", "Angle [deg]"}] :[font = output; output; inactive; preserveAspect; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = input; preserveAspect; startGroup] Plot[ fi'[t], {t, 0, T}, PlotLabel -> "Connecting Rod Angular Velocity", AxesLabel -> {"t [sec]", "[deg/sec]"}] :[font = output; output; inactive; preserveAspect; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = text; inactive; preserveAspect] Maximum and minimum values of connecting rod angular velocity. :[font = input; preserveAspect; startGroup] {fi'[T/2], fi'[0]} :[font = output; output; inactive; preserveAspect; endGroup] {34.1942057533583, -34.1942057533583} ;[o] {34.1942, -34.1942} :[font = input; preserveAspect; startGroup] Plot[ fi''[t], {t, 0, T}, PlotLabel -> "Connecting Rod Angular Acceleration", AxesLabel -> {"t [sec]", "[deg/sec^2]"}] :[font = output; output; inactive; preserveAspect; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = text; inactive; preserveAspect] Maximum and minimum values of connecting rod angular acceleration :[font = input; preserveAspect; startGroup] {fi''[T/4], fi''[3T/4]} :[font = output; output; inactive; preserveAspect; endGroup; endGroup] {3788.467538475545, -3788.467538475545} ;[o] {3788.47, -3788.47} :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] Piston Position :[font = input; preserveAspect; startGroup] Plot[ x[t], {t,0,T}, PlotLabel -> "Piston Position", AxesLabel -> {"t [sec]", "[cm]"}] :[font = output; output; inactive; preserveAspect; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = input; preserveAspect; startGroup] Plot[ x'[t], {t,0,T}, PlotLabel -> "Piston Velocity", AxesLabel -> {"t [sec]", "[cm/sec]"}] :[font = output; output; inactive; preserveAspect; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = text; inactive; preserveAspect] Minimum Piston velocity :[font = input; preserveAspect; startGroup] FindMinimum[ x'[t], {t,.01}] :[font = output; output; inactive; preserveAspect; endGroup] {-8.81696880802745, {t -> 0.01224074830705146}} ;[o] {-8.81697, {t -> 0.0122407}} :[font = text; inactive; preserveAspect] Maximum Piston velocity :[font = input; preserveAspect; startGroup] FindMinimum[ -x'[t], {t,.05}] :[font = output; output; inactive; preserveAspect; endGroup] {-8.81696880802745, {t -> 0.04775925169294854}} ;[o] {-8.81697, {t -> 0.0477593}} :[font = input; preserveAspect; startGroup] x'[0.0477593] :[font = output; output; inactive; preserveAspect; endGroup] 8.81696880788181 ;[o] 8.81697 :[font = input; preserveAspect; startGroup] Plot[ x''[t], {t,0,T}, PlotLabel -> "Piston Acceleration", AxesLabel -> {"t [sec]", "[cm/sec^2]"}] :[font = output; output; inactive; preserveAspect; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = text; inactive; preserveAspect] Find minimum piston accelerations :[font = input; preserveAspect; startGroup] x''[0] :[font = output; output; inactive; preserveAspect; endGroup] -1163.762877225956 ;[o] -1163.76 :[font = text; inactive; preserveAspect] Find maximum piston acceleration :[font = input; preserveAspect; startGroup] FindMinimum[ -x''[t], {t,.02}] :[font = output; output; inactive; preserveAspect; endGroup] {-613.2657339810624, {t -> 0.03676005757114008}} ;[o] {-613.266, {t -> 0.0367601}} :[font = input; preserveAspect; startGroup] x''[.0367601] :[font = output; output; inactive; preserveAspect; endGroup; endGroup; endGroup] 613.2657339774163 ;[o] 613.266 :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Explicit values of the Desired Functions. :[font = text; inactive; preserveAspect] The solution is also obtainable with a graphing calculator. Let f be f and q be q. Suppose that you calculate f as follows: f = p - sin-1( r/L sin(q)). IMPORTANT: Calculators return a value of arcsin between -90 and +90 degrees. Our values of interest lie between about 150 and 210 degrees. Therefore the adjustment must be made as shown. Given next are the functions to be loaded. ;[s] 13:0,0;68,1;70,2;80,3;81,4;139,5;140,6;143,7;144,8;151,9;153,10;163,11;164,12;550,-1; 13:1,11,8,Times,0,12,0,0,0;1,0,0,Symbol,0,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,0,0,Symbol,0,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,0,0,Symbol,0,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,0,0,Symbol,0,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,11,8,Times,32,12,0,0,0;1,11,8,Times,0,12,0,0,0;1,0,0,Symbol,0,12,0,0,0;1,11,8,Times,0,12,0,0,0; :[font = subsubsection; inactive; preserveAspect; startGroup] The equation describing f :[font = input; preserveAspect; startGroup] e1 = Sin[f[t]] L == Sin[q[t]] r :[font = output; output; inactive; preserveAspect; endGroup; endGroup] L*Sin[f[t]] == r*Sin[q[t]] ;[o] L Sin[f[t]] == r Sin[q[t]] :[font = subsubsection; inactive; preserveAspect; startGroup] The function describing f' :[font = input; preserveAspect; startGroup] e2 = D[e1,t] /. q'[t] -> w :[font = output; output; inactive; preserveAspect; endGroup] L*Cos[f[t]]*Derivative[1][f][t] == r*w*Cos[q[t]] ;[o] L Cos[f[t]] f'[t] == r w Cos[q[t]] :[font = input; preserveAspect; startGroup] s2 = Solve[e2,f'[t]][[1,1]] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] Derivative[1][f][t] -> (r*w*Cos[q[t]]*Sec[f[t]])/L ;[o] r w Cos[q[t]] Sec[f[t]] f'[t] -> ----------------------- L :[font = subsubsection; inactive; preserveAspect; startGroup] The function describing f'' :[font = input; preserveAspect; startGroup] e3 = D[e2,t] :[font = output; output; inactive; preserveAspect; endGroup] -(L*Sin[f[t]]*Derivative[1][f][t]^2) + L*Cos[f[t]]*Derivative[2][f][t] == -(r*w*Sin[q[t]]*Derivative[1][q][t]) ;[o] 2 -(L Sin[f[t]] f'[t] ) + L Cos[f[t]] f''[t] == -(r w Sin[q[t]] q'[t]) :[font = input; preserveAspect; startGroup] s3 = Solve[e3,f''[t]][[1,1]] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] Derivative[2][f][t] -> -((Sec[f[t]]*(-(L*Sin[f[t]]*Derivative[1][f][t]^2) + r*w*Sin[q[t]]*Derivative[1][q][t]))/L) ;[o] f''[t] -> -( 2 Sec[f[t]] (-(L Sin[f[t]] f'[t] ) + r w Sin[q[t]] q'[t]) -------------------------------------------------------) L :[font = subsubsection; inactive; preserveAspect; startGroup] The function x :[font = input; preserveAspect; startGroup] e4 = X[t] == r Cos[w t] + L Cos[ f[t] ] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] X[t] == r*Cos[t*w] + L*Cos[f[t]] ;[o] X[t] == r Cos[t w] + L Cos[f[t]] :[font = subsubsection; inactive; preserveAspect; startGroup] The function x' :[font = input; preserveAspect; startGroup] e5 = D[e4,t] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] Derivative[1][X][t] == -(r*w*Sin[t*w]) - L*Sin[f[t]]*Derivative[1][f][t] ;[o] X'[t] == -(r w Sin[t w]) - L Sin[f[t]] f'[t] :[font = subsubsection; inactive; preserveAspect; startGroup] The function x'' :[font = input; preserveAspect; startGroup] e6 = D[e5,t] :[font = output; output; inactive; preserveAspect; endGroup; endGroup; endGroup; endGroup] Derivative[2][X][t] == -(r*w^2*Cos[t*w]) - L*Cos[f[t]]*Derivative[1][f][t]^2 - L*Sin[f[t]]*Derivative[2][f][t] ;[o] 2 2 X''[t] == -(r w Cos[t w]) - L Cos[f[t]] f'[t] - L Sin[f[t]] f''[t] :[font = section; inactive; Cclosed; preserveAspect; startGroup] ISSUES IN SOLUTION :[font = text; inactive; preserveAspect; endGroup; endGroup] This problem may be solved using other notational schemes depending on the course context. In a course in dynamics, (2nd year mechanical engineers), vectors would probably be used. In a course in machinery kinematics, (3rd year mechanical engineers), the approach would be to construct loop equations in the complex plane. In either case, the results would be the same: a representation of the configuration as it evolves in time. A deep understanding of the derivation of the solution of the problem as presented here would be a valuable first step in the understanding of more complex machines. ^*)